Ahouse painter stands 3.0 m above the ground on a 5.0−m-long ladder that leans against the wall at a point 4.7 m above the ground. the painter weighs 765 n and the ladder weighs 138 n. assuming no friction between the house and the upper end of the ladder, find the force of friction that the driveway exerts on the bottom of the ladder.

fs =202.3N

Explanation:

The equilibrium equation are:

∑Fx=0  

∑Fy=0

∑M = 0  

Where:

∑Fx :algebraic sum of forces in the direction of the x-axis  

∑Fy: algebraic sum of forces in the direction of the y-axis  

∑M : Algebraic sum of moments

M = F*d  

M : moment  ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point  ( meters )

Data

W₁ =138 N  :weight of the ladder

W₂ =765 N : weight of the house painter

L = 5 m : ladder length

h =  4.7 m : ladder height

µ = 0 : coefficient of friction between the ladder and the wall

θ  : angle that makes  the  ladder  with the floor

sinθ = h/L = 4.7 m / 5 m

θ =sin⁻¹(4.7 / 5)

θ = 70.05°

Forces acting on the ladder

W₁ = 138 N :Weight  of the ladder (vertical downward)

W₂ =765 N  : Weight of the fire painter(vertical downward)  

FN₁ :Normal force that the floor exerts on the ladder (vertical upward)  (point A)

fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)  

FN₂ :  Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)

Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)

d₁ =2.5 *cos 70.05°   = 0.853 m: Horizontal distance from W₁ to the point A

d₂ = 3 / tan 70.05°  = 1.089 m  : Horizontal distance from W₂ to the point A

d₃ = 4.7 m : Vertical distance from FN₂ to the point A

Equilibrium equation of the moments at the point A  (contact point of the ladder with the floor)  

∑MA = 0  

FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0

FN₂(d₃) = W₁(d₁) + W₂(d₂)

FN₂( 4.7) = ( 138 )(  0.853) + (765)( 1.089 )

FN₂( 4.7) = 117.714 + 833.085

FN₂ = (950.799) / ( 4.7)  

FN₂ =202.3 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Equilibrium equation of forces in x-direction

∑Fx=0

FN₂ - fs = 0

202.3 N - fs = 0

202.3 N =  fs

fs  = 202.3 N : force of friction that the driveway exerts on the bottom of the ladder

measuring things 1 litre of water

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.